Equivalence of categories preserves subobject classifiers.

Solution 1:

Since you've asked for an answer using only the concepts in the question, I'll do my best to give such a proof. However, before I do, I want to say that Hanno's suggestion in the comments is definitely the best way to prove this.

Also your proofs have some issues. I won't point out all the differences, because this answer will be long enough as is, but please take note of them.

If $F:C\to D$ is an equivalence of categories, $\alpha : 1\to \Omega\in C$ is a subobject classifier of $C$, then $F\alpha : F1\to F\Omega$ is a subobject classifier of $D$.

Proof:

As you've noted, we should prove that $F1$ is a terminal object of $D$, and that $F\alpha$ remains a monomorphism.

$F1$ is terminal: Let $X\in D$, since $F$ is essentially surjective, there exists $A\in C$ such that $FA\cong X$. Then $$D(X,F1)\cong D(FA,F1)\cong C(A,1)\cong \{*\}.$$ Thus $F1$ is terminal in $D$, since all hom sets $D(X,F1)$ are one element sets.

$F$ preserves (and reflects) monomorphisms: Let's rephrase the property of being a monomorphism. $i:X\to Y$ is a monomorphism in a category $C$ if for all objects $Z\in C$, $g,h:Z\to X$, $ig=ih \implies g=h$. In other words, the map $i_* : C(Z,X)\to C(Z,Y)$ is injective for all objects $Z\in C$.

Now suppose $i:X\to Y$ is a monomorphism in $C$, so $i_*:C(Z,X)\to C(Z,Y)$ is injective. Then we have the diagram $$\require{AMScd} \begin{CD} C(Z,X) @>i\circ ->> C(Z,Y)\\ @V\simeq VFV @V\simeq VFV \\ D(FZ,FX) @>Fi\circ - >> D(FZ,FY), \\ \end{CD} $$ so $Fi_*$ is injective a map from $D(FZ,FX)$ to $D(FZ,FY)$ for all $Z\in C$. This is not quite what we want though, but it's good enough. For any object $A\in D$, we can find $Z\in C$ such that $\phi : A\xrightarrow{\cong} FZ$. Then the composite $$D(A,FX)\newcommand\toby\xrightarrow \toby{-\circ \phi^{-1}} D(FZ,FX) \overset{Fi\circ}{\hookrightarrow} D(FZ, FY) \toby{-\circ\phi} D(A,FY)$$ is injective, and reduces to just the map $f\mapsto Fi\circ f \circ \phi^{-1}\circ \phi = Fi\circ f$. Thus $Fi$ is a monomorphism in $D$.

In particular, $F\alpha : F1\to F\Omega$ is a monomorphism.

Note also that if $Fi$ is a monomorphism in $D$, then by the commutative square above, we see that $i$ must have been a monomorphism in $C$.

$F\alpha : F1\to F\Omega$ is a subobject classifier: Let $i : A\to X$ be a monomorphism in $D$. Choose $B,Y\in C$ such that $A\cong FB$ and $X\cong FY$. Let $J : FB\to FY$ be the conjugate monomorphism of $i$. Fixing these isomorphisms, then there is a unique morphism $\varphi : X\to F\Omega$ making $$ \begin{CD} A @>>> F1 \\ @VVi V @VV F\alpha V\\ X @>\varphi>> F\Omega \\ \end{CD} $$ a pullback square if and only if there is a unique morphism $\Phi : FY\to F\Omega$ making $$ \begin{CD} FB @>>> F1 \\ @VVJ V @VV F\alpha V\\ FY @>\Phi>> F\Omega \\ \end{CD} $$ a pullback square. Thus it suffices to prove that there is such a unique morphism $\Phi$.

Now since $F:C(B,Y)\to D(FB,FY)$ is a bijection, there is some morphism $j:B\to Y$ such that $J=Fj$. By the note in the previous section, $j$ is a monomorphism. Thus since $\alpha$ is a subobject classifier in $C$, we have a unique morphism $\psi:Y\to \Omega$ such that the following diagram is a pullback square: $$ \begin{CD} B @>>> 1 \\ @VVj V @VV \alpha V\\ Y @>\psi>> \Omega. \\ \end{CD} $$

It just remains to prove that $\Phi = F\psi$ has the desired property. Certainly the square commutes, and if $\Phi$ were a morphism making the square commute, then since $F$ induces a bijection on hom sets, we could find a map $\psi$ such that $F\psi=\Phi$ making the corresponding square in $C$ commute. Thus it suffices to prove that $$ \begin{CD} B @>>> 1 \\ @VVj V @VV \alpha V\\ Y @>\psi>> \Omega \\ \end{CD} $$ is a pullback square if and only if $$ \begin{CD} FB @>>> F1 \\ @VVFj V @VV F\alpha V\\ FY @>F\psi >> F\Omega \\ \end{CD} $$ is a pullback square.

This is the same idea as before when we were proving $F1$ was terminal, or that $Fi$ is a monomorphism if and only if $i$ is a monomorphism. In fact, the general fact is that if $F$ is an equivalence of categories, then it both preserves and reflects all limits and colimits. I'll prove that $F$ preserves and reflects all limits, since the general language of cones to a diagram is more convenient than talking about pullbacks specifically.

$F$ preserves and reflects all limits:

Lemma: Let $I:A\to C$ be a diagram in $C$. Let $F:C\to D$ be an equivalence of categories. Let $(X,\newcommand\set[1]{\left\{{#1}\right\}}\set{f_a})$ be a cone to $I$. $(X,\set{f_a})$ is a limit cone if and only if $(FX,\set{Ff_a})$ is a limit cone to $FI$.

Proof

First suppose $(FX,\set{Ff_a})$ is a limit cone. Then given a cone $(Y,\set{g_a})$ to $I$, the cone $(FY,\set{Fg_a})$ induces a unique map $\alpha_0 : FY\to FX$ such that $Ff_a \circ \alpha_0 = Fg_a$. Let $\alpha : Y\to X$ be the corresponding map, so that $F\alpha = F\alpha_0$. Then we have that $f_a \circ \alpha = g_a$, and $\alpha$ is unique, since if $\tilde{\alpha}$ also had this property, then $F\tilde{\alpha} : FY\to FX$ would satisfy the same property as $\alpha_0$. So by uniqueness, $F\tilde{\alpha} = \alpha_0= F\alpha$. Hence $\tilde{\alpha}=\alpha$. Thus $(X,\set{f_a})$ is a limit cone. Therefore $F$ reflects limits.

Now suppose $(X,\set{f_a})$ were a limit cone. If $(Z,\set{G_a})$ is a cone to $FI$ in $D$, then choose $Y\in C$ with $Z\cong FY$. Then we have a corresponding cone $(FY,\set{G_a'})$. As usual, we can now reflect the morphisms into $C$ to get $g_a : Y\to Ia$ such that $(Y,\set{g_a})$ is a cone to $I$ in $C$. Then the unique morphism $Y\to X$ induced by this cone becomes a unique morphism $FY\to FX$, and thus a unique morphism $Z\to FX$. Thus $(FX,\set{Ff_a})$ is a limit cone in $D$.

This completes the proof of both the lemma, and the original claim. $\blacksquare$